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Emergency probability help needed s1 exam tomorrow!!!

(ii) Show that the events ‘the score on the first die is even’ and ‘the product of the scores on the two
dice is less than 10’ are not independent.
ON THIS PAPER
http://mei.org.uk/files/papers/2011_Jan_s1.pdf
I just dont get it.
Work out both probabilities, probabilities of both occurring, then times the two probabilities together and prove it doesn't equal the probability of both occurring.
Revision notes:

For dependent events (one event that depend on the other) : P(AnB) = P(A) x P(B|A) (probability of B given A has occurred).

For independent events (events that occur without influence): P(AnB)=P(A) x P(B)
If A and B are independent, P(B|A) = P(B|A') = P(B)

If they are dependent they don't equal as it depends on the outcome of A.

An example of independent would be the probability of me eating cereal or toast for breakfast and the probability of me passing the exam. They have no dependency on each other at all.

A dependent event would be the probability I forgot my calculator and the probability I passed the exam. The calculator forgetting would stress me out, id have to find one and I'd just panick too much, lowering my chances, there P(B) (me passing) would change in occurance to P(A) (me forgetting my calculator.)
(edited 6 years ago)
Original post by ExoIceCream99
Work out both probabilities, probabilities of both occurring, then times the two probabilities together and prove it doesn't equal the probability of both occurring.


thank you but i dont know how to get the probability of both occurring p(AnB)
can you please explain it to me would be much appreciated.
I know how to get the first P(A)*P(B)
Original post by AsiyaAden9879
thank you but i dont know how to get the probability of both occurring p(AnB)
can you please explain it to me would be much appreciated.
I know how to get the first P(A)*P(B)


Look at my second comment
Original post by ExoIceCream99
Look at my second comment


but how do i work out BIA don't i need p(AnB)
sorry man im so bad at probability. Could you please just give a solution i have strong feeling this is gnna come up tmrw
Original post by AsiyaAden9879
but how do i work out BIA don't i need p(AnB)
sorry man im so bad at probability. Could you please just give a solution i have strong feeling this is gnna come up tmrw


P(B|A) = probability of B and A occurring / probability of A
Original post by ExoIceCream99
P(B|A) = probability of B and A occurring / probability of A


is the probability of b and a occurring the P(B)*P(A)
So if the probability of me having cereal is 0.8 P(A) and P(B) me passing is 0.6

P(B|A) = 0.8 x 0.6 / 0.8
P(B|A') = 0.2 x 0.6 / 0.2

If they equal the same (which in this case they both equal 0.6) then they are independent events.
Original post by ExoIceCream99
So if the probability of me having cereal is 0.8 P(A) and P(B) me passing is 0.6

P(B|A) = 0.8 x 0.6 / 0.8
P(B|A':wink: = 0.2 x 0.6 / 0.2

If they equal the same (which in this case they both equal 0.6) then they are independent events.


Alright thank you so much good luck
for tomorrow.
Original post by AsiyaAden9879
Alright thank you so much good luck
for tomorrow.


You too!!!

You'll be fine :smile:
Original post by ExoIceCream99
You too!!!

You'll be fine :smile:


Thanks I hope so!

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